Let $a^3-b^3=n$ where $a>b$ are positive rational numbers. We solve the equation $x^3+y^3=a^3-b^3$ The first solution is $x=a,y=-b$ The second solution is $x=\frac{a(a^3-2b^3)}{a^3+b^3};\;y=\frac{b(2a^3-b^3)}{a^3+b^3}$ and if $\left(\frac a b\right)^3 > 2$, then $x>0,y>0$ Тhe third solution is $x=\frac{a^9+3a^6b^3-6a^3b^6+b^9}{3ab(a^6-a^3b^3+b^6)};\;y=-\frac{a^9-6a^6b^3+3a^3b^6+b^9}{3ab(a^6-a^3b^3+b^6)}$ and if $\left(\frac a b\right)^3 < 2$, then $x>0,y>0$

Pal702004 wrote: Тhe third solution is $x=\frac{a^9+3a^6b^3-6a^3b^6+b^9}{3ab(a^6-a^3b^3+b^6)};\;y=-\frac{a^9-6a^6b^3+3a^3b^6+b^9}{3ab(a^6-a^3b^3+b^6)}$ and if $\left(\frac a b\right)^3 < 2$, then $x>0,y>0$ No, sorry if $a\approx b$ then $x<0$

Ок, let's work only with the first solution if $(x_k,y_k)$ is a solution and $y_k<0$, then $x_{k+1}=\frac{y_k(2x_k^3-y_k^3)}{x_k^3+y_k^3};\;\; y_{k+1}=\frac{x_k(x_k^3-2y_k^3)}{x_k^3+y_k^3}$ is also solution. And if $\frac{x_k}{|y_k|}>\sqrt[3] 2$ then $x_{k+1}>0$ and $y_{k+1}>0$ If $\frac{x_k}{|y_k|}<\sqrt[3] 2$, then $y_{k+1}$ is negative, but $\frac{x_{k+1}}{|y_{k+1}|}>\frac{x_{k}}{|y_{k}|}$ Let $\frac{x_{k}}{|y_{k}|}=z$ $\frac{x_{k+1}}{|y_{k+1}|}=\frac{2z^3-1}{z(2-z^3)}>z\; \; \forall z \in (1;\sqrt[3] 2)$ $\Rightarrow \exists n:\;\; \frac{x_{n}}{|y_{n}|}>\sqrt[3] 2$ and $x_{n+1}>0 $ and $y_{n+1}>0$

Isn't this problem 'too' easy for TST ?

I don't think the solution in post #02 is obvious at all Could you explain your motivation if you think this is too easy for a TST

Groupsolved with biomathematics, pluto1708 We want to solutions $(c,d)$ of the $a^3-b^3=c^3-d^3$ from $(a,b)$ with a negative $d$. First, we simply try to generate new roots. Observe that if we have some solution $x,y$ of the equation $x^3-y^3=(a^3-b^3)(a^3+b^3)^3$ then $(c,d)=(\frac{x}{a^3+b^3},\frac{y}{a^3+b^3})$ is a solution of $a^3-b^3=c^3-d^3$. Now, observe that $(a^3-b^3)(a^3+b^3)^3=(a-\omega b)(a+\omega b)^3(a-\omega^2 b)(a+\omega^2 b)^3(a-b)(a+b)^3$. Now, we expand to get \[(a^3-b^3)(a^3+b^3)=(a^4-2ab^3-\omega(b^4-2a^3b))(a^4-2ab^3-\omega^2(b^4-2a^3b))(a^4-2ab^3-(b^4-2a^3b))\]Thus, $(a^3-b^3)(a^3+b^3)^3=(a^4-2ab^3)^3-(b^4-2a^3b)^3=(2a^3b-b^4)^3-(2ab^3-a^4)$. Thus, we get that $a^3-b^3=c^3-d^3$ has solution $(c,d)=(\frac{2a^3b-b^4}{a^3+b^3},\frac{2ab^3-a^4}{a^3+b^3})$. So, let $f(a,b)=(\frac{2a^3b-b^4}{a^3+b^3},\frac{2ab^3-a^4}{a^3+b^3})$. Now, observe that in our problem, $a^3-b^3$ is positive and thus, $a>b$ and in our new solution as well, we get $\frac{2a^3b-b^4}{a^3+b^3}>0$. Now, if $\frac{2ab^3-a^4}{a^3+b^3}$ is negative then we are done. But, if it is positive, this implies that $a>b$ and $2^{\frac{1}{3}}>a$. Observe that this $\implies \frac{\frac{2a^3b-b^4}{a^3+b^3}}{\frac{2ab^3-a^4}{a^3+b^3}}\ge \frac{a^3b}{ab^3}\ge (\frac{a}{b})^2 $ Now, pick $k$ such that $(\frac{a}{b})^{2^k}>2^{1/3}$. Now, if $f(a,b),\cdots f^{k}(a,b)$ don't work then we have $f^k(a,b)=(c,d)$ and $\frac{c}{d}\ge (\frac{a}{b})^{2^k}\ge 2^{1/3}$ but then $f^{k+1}(a,b)$ is a solution of the original equation.

Mr.C wrote: Let $n$ be an integer greater than $1$ such that $n$ could be represented as a sum of the cubes of two rational numbers, prove that $n$ is also the sum of the cubes of two non-negative rational numbers. Proposed by Navid Safaei Solved with AaronAKJ, alexiaslexia, Sugiyem. We will be proving the following statement instead. Generalized Version wrote: For any $z, w \in \mathbb{Q}^+$ such that $z^3 - w^3 > 0$, then there exists $x,y \in \mathbb{Q}^+$ such that \[ x^3 + y^3 = z^3 - w^3 \] Part 01. The "ingenious" construction. Proof. Let $\omega$ be the $3^{rd}$ root of unity. Notice that \begin{align*} (z^3 - w^3)(z^3 + w^3)^3 &= ((z - w)(z + w)^3) \cdot ((z - w \omega)(z + w \omega)^3) \cdot ((z - w \omega^2)(z + w \omega^2)^3) \\ &= (z^4 - 2zw^3 + (2z^3 w - w^4))(z^4 - 2zw^3 + (2z^3 w - w^4) \omega)(z^4 - 2zw^3 + (2z^3 w - w^4) \omega^2) \end{align*}Therefore, we can always pick \[ (x,y) = \left( \frac{z^4 - 2zw^3}{w^3 + z^3} , \frac{2z^3 w - w^4}{w^3 + z^3} \right) \]if $\frac{z}{w} \ge \sqrt[3]{2}$. Part 02. The "Recursion" Trick. For a fixed pair $(z,w)$. Take $f(z) = w(2z^3 - w^3), f(w) = z(2w^3 - z^3)$, we could verify easily that \[ \frac{f(z)}{f(w)} > \left( \frac{z}{w} \right)^2. \]Since $\frac{z}{w} > 1$, we have \[ \lim_{n \to \infty} \frac{f^n(z)}{f^n(w)} \ge \lim_{n \to \infty} \left( \frac{z}{w} \right)^{2^n} \]and it is clear that RHS diverges. Therefore, we can pick $(z,w) \mapsto (f^n(z), f^n(w))$ such that $\frac{f^n(z)}{f^n(w)} > \sqrt[3]{2}$. Hence, we are done.

Another motivation for the construction is the theory of elliptic curves: the main idea is that we pick two rational points $A$ and $B$ in a cubic curve and compute the line passing through them; the third intersection is a rational point $A*B$. Then we reflect across a convenient line (corresponding actually to using a line at infinity) and find another rational point $A+B$. The plus sign is no coincidence: this operation defines a group of points. We can define $A+A=2A$ by using a tangent line through $A$, which is what we do in this problem. Let $C\colon x^3-y^3=n$ be such a curve and $P=(a,b)$ one of its points. Then draw the tangent $t\colon y=\frac{a^2}{b^2}x - \frac n{b^2}$ (use some calculus to find the equation: $x^3-y^3=n\implies 3x^2\,dx - 3y^2\,dy=0$) to $C$ at $(a,b)$ and find the other intersection $P^*$ between $t$ and $C$. Finally, find the reflection $2P$ of $P^*$ with respect to the line $x+y=0$. The intersection can be found by solving the cubic $x^3-\left(\frac{a^2}{b^2}x-\frac n{b^2}\right)^3 = n$, which has leading coefficient $1-\frac{a^6}{b^6} = \frac{b^6-a^6}{b^6} = -\frac{n(a^3+b^3)}{b^6}$ and independent term $\frac{n^3}{b^6}-n = -\frac{n(b^6-n^2)}{b^6} = -\frac{na^3(2b^3-a^3)}{b^6}$. We know that $a$ is a double root, so the other root, by Vieta's relations, is $-\frac{a^3(2b^3-a^3)}{a^2(a^3+b^3)} = -\frac{a(2b^3-a^3)}{a^3+b^3}$. Plugging in $y=\frac{a^2}{b^2}x - \frac n{b^2}$ we find $y = -\frac{b(2a^3-b^3)}{a^3+b^3}$. Then the reflection is $\left(\frac{b(2a^3-b^3)}{a^3+b^3},\frac{a(2b^3-a^3)}{a^3+b^3}\right)$. The following steps are the same as the other solutions presented here, which is computing $2^kP$.

Similar problem: Sum of 5th Powers of 2 Nonnegative Rational Numbers? This deleted post asked: Is the following statement also true? wrote: \(n\in\mathbb{N}\) can be written in the form that sum of the sursolids of two nonnegative rational numbers iff \(n\) is a sum of 2 rational sursolids.

Nice problem! Sketch: First reduce to solving $u^3-v^3=(a^3-b^3)(a^3+b^3)^3$. By complex numbers trick, we get the identity $(x^3-y^3)(x^3+y^3)^3=(2x^3y-y^4)^3-(2xy^3-x^4)^3$, if $\left(\frac{2xy^3-x^4}{x^3+y^3}\right) < 0$ then we're done, else if the idea is to notice that the ratio of the two numbers is $\geq \frac{a^2}{b^2}$ then raising to sufficiently large power works